الأسبوع 1 · قُوى القُوى والمعادلات الأسية: قوانين الأسس للأعداد الحقيقية والأسس الصحيحة والمعادلة الأسية من الصورة أس المتغير يساوي عددًا · 3 دروس
2
Simplifying expressions with powers in x, y and n, then finding their value
Mathematics Prep 3 Term 1, Unit 1, Lessons 1–2
GoalSimplify expressions with positive, zero and negative exponents, say which letters must not be zero, and find the value for a given number.
New words
zero exponent · a⁰ = 1 for any a ≠ 0negative exponent · a⁻ⁿ = 1 ÷ aⁿ, the reciprocal of aⁿsimplest form · every letter appears once, with a positive exponentvalue of an expression · the number you get when the letters are replaced by given numbers
Explanation
You already know aᵐ × aⁿ = aᵐ⁺ⁿ, aᵐ ÷ aⁿ = aᵐ⁻ⁿ, a⁰ = 1 and a⁻ⁿ = 1 ÷ aⁿ. Now combine them with the three laws of Lesson 1. Work in this order: first remove brackets by giving every factor the outside exponent, then collect the same letters by adding exponents (multiplying) or subtracting them (dividing), and last write negative exponents as fractions. A letter that is in a denominator, or has a negative exponent, must not be zero, so say it: x ≠ 0. To find a value, simplify first and then replace the letters; this is shorter and safer than putting the numbers into the long form.
Rule
Example
a⁰ = 1 (a ≠ 0)
7⁰ = 1, (−x)⁰ = 1
a⁻ⁿ = 1 ÷ aⁿ
2⁻³ = 1 ÷ 8, x⁻² = 1 ÷ x²
aᵐ × aⁿ = aᵐ⁺ⁿ
x² × x⁻⁵ = x⁻³
aᵐ ÷ aⁿ = aᵐ⁻ⁿ
y³ ÷ y⁷ = y⁻⁴
Worked examples
Simplify (2x²y⁻¹)³ ÷ (4x⁴), with x ≠ 0 and y ≠ 0. First the bracket: (2x²y⁻¹)³ = 8x⁶y⁻³. Divide by 4x⁴: the numbers give 8 ÷ 4 = 2 and the x's give x⁶⁻⁴ = x². So the result is 2x²y⁻³ = 2x² ÷ y³.
Find the value of (x²)³ × x⁻⁵ ÷ x⁻² when x = −2. Simplify first: x⁶ × x⁻⁵ = x¹, and x¹ ÷ x⁻² = x¹⁺² = x³. For x = −2 the value is (−2)³ = −8.
A school hall floor is a square of side 5x² cm and it is covered with square tiles of side x cm. How many tiles are needed? Floor area = (5x²)² = 25x⁴ cm². One tile = x² cm². Number of tiles = 25x⁴ ÷ x² = 25x². With x = 20 the tile is 20 cm, the hall is 2000 cm = 20 m wide, and 25 × 20² = 10 000 tiles are needed.
Class activity
Halving game. Write 2⁵, 2⁴, 2³, 2², 2¹ with their values 32, 16, 8, 4, 2. Each step divides by 2. Keep going with 2⁰, 2⁻¹, 2⁻², 2⁻³. What values do you get? Does this explain why 2⁰ = 1?
Practice
1
Simplify x⁵ × x⁻⁷ (x ≠ 0).
x⁻² = 1 ÷ x².
2
Simplify (3a⁻²b)² (a ≠ 0).
9a⁻⁴b² = 9b² ÷ a⁴.
3
Find (2⁻¹)³ × 2⁵.
(2⁻¹)³ = 2⁻³, so 2⁻³ × 2⁵ = 2² = 4.
4
A cube has side x² cm. Find its volume, and its value for x = 3.
V = (x²)³ = x⁶ cm³. For x = 3: 3⁶ = 729 cm³.
5
A cylinder has radius 2x cm and height x² cm. Write its volume πr²h in simplest form.
